Question

Four containers of $2L$ capacity contains dinitrogen as describe below. Which one contains a maximum number of molecules?

• A. $2.5$ g-molecule of $N_2$
• B. $4$ g-atom of nitrogen
• C. $3.01\times {10}^{24}N$ atoms
• D. $82$ g of dinitrogen.

Answer 1

Answer: (C)

$3.01\times {10}^{24}N$ atoms

For $A$:- No. of molecules= $2.5\times 6.022\times {10}^{23}$

=$2.5\times N_A$ molecules

For $B$:- No. of atoms=$\frac{4}{14}\times 2\times N_A$

=$0.57N_A$ molecules

For $C$:- No. of atoms=$\frac{3.01\times {10}^{24}}{6.022\times {10}^{23}}{N}_A=5N_A$ atoms

For $D$:- No. of molecules=$\frac{82}{28}\times N_A=2.92N_A$ molecules

Therefore, $3.01\times {10}^{24}N$ contains maximum number of molecules